Section - 2 TANGENTS · Mathematics / Ellipse As in circles and parabolas, the equation of a...

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Mathematics / Ellipse

As in circles and parabolas, the equation of a tangent to a given ellipse can take various different forms, all of which

we discuss in this section. We will use the ellipse 2 2

2 2 1x ya b

as our standard throughout this discussion.

Consider the ellipse2 2

2 2( , ) : 1x yS x ya b

and a point P(x1, y1) lying on this ellipse.

Thus,2 21 12 2 1x y

a b...(1)

The slope Tm of the tangent at P(x1, y1) can be obtained by evaluating the derivative ofthe curve at P. For this purpose, we differentiate the equation of the ellipse :

2 2

2 2 0x y dya b dx

2

2

dy b xdx a y

1 1

21

2( , ) 1

TP x y

dy b xmdx a y

The equation of the tangent can now be obtained using point-slope form :2

11 12

1

( )b xy y x xa y

2 21 1 1 1

2 2 2 2

xx yy x ya b a b

...(2)

Using (1), the RHS in (2) is 1 so that the equation of the ellipse is

1 12 2 1xx yy

a b

The equation obtained for the tangent can be, as in the case of circles and parabolas,written concisely in the form

1 1( , ) 0T x y

Section - 2 TANGENTS



If the point P is specified in parametric form instead of cartesian form, we simply substitute

1 1cos , sinx a y b in the equation of the tangent obtained above. Thus, theequation in this case is

cos sin 1x ya b

In example-15, we proved that any line of the form

2 2 2y mx a m b

is a tangent to the ellipse 2 2

2 2 1,x ya b

whatever the value of m may be.

As an exercise, show that this tangent touches the ellipse at the point

2 2

2 2 2 2 2 2,a m b

a m b a m b

Also show that from any point P, in general two tangents (real or imaginary)can bedrawn to the ellipse (use the approach followed in Circles)

Tangents drawn at 1( )A and 2( )B on the ellipse 2 2

2 2 1x ya b

intersect in P. Find the coordinates of P.

The equations of the tangents at A and B, using the parametric form for the tangent, are

1 1cos sin 1x ya b ...(1)

2 2cos sin 1x ya b

...(2)

2 1(1) cos (2) cos gives

1 2 2 1sin( ) cos cosyb

2 1

1 2

(cos cos )sin( )

y b

1 2

1 2

sin2

cos2

b

Example – 16



Similarly,

2 1(1) sin (2) sin gives

1 2

1 2

cos2

cos2

ax

Thus, the coordinates of the point of intersection P are

1 2 1 2

1 2 1 2

cos sin2 2,

cos cos2 2

a bP

Find the locus of a moving point P such that the two tangents drawn from it to an ellipse are perpendicular.

Let the equation of the ellipse be 2 2

2 2 1x ya b

and P be the point (h, k).

Any tangent to this ellipse is of the form

2 2 2y mx a m b

If this passes through P(h, k), we have

2 2 2k mh a m b2 2 2 2( )k mh a m b

2 2 2 2 2( ) 2 0h a m hkm k b

As expected, we obtain a quadratic in m which will give us two roots, say m1 and m2. Since thetangents through P are perpendicular, we have

1 2 1m m

2 2

2 2 1k bh a

2 2 2 2h k a bThus, the locus of P is

2 2 2 2x y a b

Example – 17 DIRECTOR CIRCLE



which is a circle and is termed the of the ellipse. From point on the DirectorCircle of an ellipse, the two tangents drawn to the ellipse are perpendicular.

y

xFrom any point on the director circle of an ellipse, the two tangents drawn to the ellipse are perpendicular.

P

Prove that the product of the perpendiculars from the foci upon any tangent to the ellipse 2 2

2 2 1x ya b

is 2 .b

We can assume an arbitrary tangent to this ellipse to be

2 2 2y mx a m b ...(1)

The perpendicular distances of the two foci, 1( ,0)F ae and 2 ( ,0)F ae from the line given by (1) are

2 2 2

1 21mae a m bd

m2 2 2

2 21mae a m bd

mWe thus have,

2 2 2 2 2 2

1 2 21a m b a m ed d

m

2 2 2 2

2

(1 )1

a m e bm

2 2 2

21m b b

m 2b

Example – 18



Find the locus of the foot of the perpendicular drawn from the origin upon any tangent to the ellipse 2 2

2 2 1.x ya b

We can assume an arbitrary tangent to this ellipse as

2 2 2y mx a m b

Let the foot of perpendicular from the origin upon this tangent be P(h, k):y

P h,k( )

S

O x

Thus,

2 2 2k mh a m b ...(1)

Also,OP SP ( the tangent)

1k mh

hmk ...(2)

Using (2) in (1), we obtain a relation between h and k as

2 2 2 2 2 2 2( )h k a h b k

Thus, the required locus of P is

2 2 2 2 2 2 2( )x y a x b y

Prove that the portion of the tangent to any ellipse intercepted between the curve and a directrix subtends a rightangle at the focus.

Example – 19

Example – 20



The following diagram makes the phrase “corresponding focus” clear :

y

P

xF1

T1

F2

T2

We need to prove that

PF PF T1 1 2 2T = = 2

Let the equation of the ellipse be 2 2

2 2 1x ya b

and assume P to be ( cos , sin ).a b F1 and F2 are

(–ae, 0) and (ae, 0) respectively.The equation of the tangent at P is

cos sin 1x ya b

T1 and T2 can now be evaluated since we know their x-coordinates as a

e and ae respectively.

1( cos ):

sina b eT x y

e e

1( cos ),

sina b eT

e eSimilarly,

2( cos ),

sina b eTe e

Now we evaluate the appropriate slopes :

1 1 2

( cos )( cos )sin

sin (1 )FT

b eb eem a a eae

e

1

sin sincos ( cos )PFb bm

a ae a e

1 1 1

2 2

2 2 2 1(1 )FT PF

b bm ma e b

which implies 1 1 1FT PF

Similarly, we can show that 2 2 2.F T PF

Thus, the two intercepts subtend right angles at their foci.



Common tangents are drawn to the parabola 2 4y x and the ellipse 2 2

116 6x y

touching the parabola at A and

B and the ellipse at C and D. What is the area of the quadrilateral ABDC ?

y

O x

A

C

EP

D

y = x2 4

Q

B

An approximate figure showing the common tangents and intersecting in (which will lie on the axis due to the symmetry of the problem)

AC BD E

Any tangent to the parabola 2 4y x can be written in the form

1y mxm

This line touches the ellipse if the condition for tangency, 2 2 2 2c a m b is satisfied, i.e. if

22

1 10 6mm

giving1

2 2m

Thus, the two tangents AE and BE are

1 2 22 2

y x

which evidently intersect at ( 4,0).E

Example – 21



The point of contact for the parabola 2 4y ax is given by 2

2, .a am m Thus A and B have the

coordinates (8, 4 2) so that

8 2AB

The point of contact for the ellipse will be 2 2

2 2 2 2 2 2, .a m b

a m b a m b

Thus, C and D will have the coordinates 32,2 so that

3 2CDFinally, PQ can now easily be seen to be 8 + 2 = 10. The area of quadrilateral ABDC (which isactually a trapezium) is

1 ( )2

AB CD PQ

55 2 sq units

Prove that in any ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of theellipse to point of contact meet on the corresponding directrix.

Let the ellipse be 2 2

2 2 1x ya b

and let a tangent be drawn to it at an arbitrary point ( cos , sin )P a b as

shown :y

xO

P

F

T

Q

We need to show that the perpendicular from F onto this tangent, i.e., FT, and the line joining thecentre to the point of contact, i.e. OP intersect on the corresponding directrix; in other words, we

need to show that the x-coordinate of Q as in the figure above is .axe

Example – 22



The equation of the tangent at P is cos sin 1.x ya b

The slope of this tangent is cotTbm

aTherefore, the slope of FT is

tanFTamb

The equation of FT is

: tan ( )aFT y x aeb ...(1)

The equation of OP is simply

: tanbOP y xa ...(2)

Comparing (1) and (2) gives ,ax e which proves the stated assertion.

Find the coordinates of all the points on the ellipse 2 2

2 2 1x ya b

for which the area of PON is the maximum

where O is the origin and N is the foot of the perpendicular from O to the tangent at P.

We can assume the point P to be ( cos , sin )a b so that the tangent at P has the equation

cos sin 1x ya b ...(1)

O

P a , b ( cos sin )N

y

x

To evaluate the area of ,PON we first need the coordinates of the point N. The equation of ON is

0 tan ( 0)ay xb

sin cos 0ax by ...(2)

Example – 23



The intersection of (1) and (2) gives us the coordinates of N as

2 2

2 2 2 2 2 2 2 2

cos sin,sin cos sin cos

ab a bNa b a b

The length PN can now be evaluated using the distance formula :

2 2

2 2 2 2

( ) sin cossin cos

a bPNa b

The length ON is simply the perpendicular distance of O from the tangent at P given by (1) :

2 2 2 2sin cosabON

a b

Thus, the area of OPN is

12

PN ON

2 2

2 2 2 2

( ) sin cos2 sin cos

ab a ba b

2 2 12 tan cot

a ba bb a

The expression tan cota bb a is of the form

1yy whose minimum magnitude is 2, when

tan 1ab

tan ba

2 2 2 2sin ,cosb a

a b a b

When this minimum is achieved, is maximum. Thus, the possible coordinates of P for which area( )OPN is maximum are

2 2

2 2 2 2,a bP

a b a b

As might have been expected from symmetry, there are four such possible points on the ellipse.



A straight line L touches the ellipse 2 2

2 2 1x ya b

and the circle 2 2 2x y r (where b < r < a). A focal chord of

the ellipse parallel to L meets the circle in A and B. Find the length of AB.

L is a common tangent to the ellipse and the circle. We can assume the equation of L to be (using theform of an arbitrary tangent to an ellipse) :

2 2 2y mx a m b

x

y

A

BF1

N

L

O

Since L is a tangent to the circle too, its distance from (0, 0) must equal r. Thus,

2 2 2

21a m b r

m

2 2

2 2

r bma r

...(1)

The equation of AB( which passes through 1( , 0))F ae can now be written as

0 ( )y m x aemx y mae ...(2)

To evaluate the length AB, one alternative is to find the intercept that the circle 2 2 2x y r makes onthe line AB given by (2).

However, a speedier approach would be to use the Pythagoras theorem in .OAN OA is simply theradius r. ON is the perpendicular distance of O from the line given by (2). Thus,

2 1maeONm

Example – 24



Finally, we have

2AB AN

2 22 OA ON

2 2 2

222

1m a erm

2 2 2 2 2 2

22 2 2 2

( )2( )

r b a e a rra b a b

= 2 b

The length of the chord AB is equal to 2b, the same as the minor axis of the ellipse.

Find the angle of intersection of the ellipse 2

2 2 1x ya b

and the circle 2 2 .x y ab

The semi-major and semi-minor axis of the ellipse are of lengths a and b respectively whereas theradius of the circle is .ab Note that

b ab a

Thus, the circle will intersect (symmetrically) the ellipse in four points.

x

y

PWe need to find

Example – 25



Consider any point of intersection, say P, the one in the first quadrant. The coordinates of P can beassumed to be ( cos , sin ).a b Since P also lies on the circle, we have

2 2 2 2cos sina b ab

2 2 2( )cos ( )a b b a b

2

2

cos

sin

ba b

aa b

...(1)

At P, the tangent to the ellipse has the slope

32

cotEb bm

a a (Using (1))

while the tangent to the circle has the slope

12

cotCa am

b b (Again, using (1))

Thus, the angle of intersection is given by

312 2

tan1 1

C E

C E

a bm m a bb a

bm m aba

1tan a bab



Prove that the tangents at the extremities of the latus rectum of an ellipse intersect on the correspondingdirectrix.

Let P and Q be points on the ellipse 2 2

2 2 1x ya b

whose eccentric angles differ by .2 Tangents at P

and Q intersect at R. What is the locus of R ?

Prove that the locus of the mid-point of the portion of the tangent to the ellipse 2 2

2 2 1x ya b

intercepted

between the coordinate axes is 2 2 2 2 2 24 .b x a y x y

We generalise Question - 2 (above) in this question. What is the locus of the point of intersection of

tangents to the ellipse 2 2

2 2 1x ya b

at points whose eccentric angles differ by a constant 2 .

Find the locus of the foot of perpendicular on any tangent to the ellipse 2 2

2 2 1x ya b

from either of its

foci.

A family of ellipses have the same major axis, but different minor axis. Prove that the tangents at theend-points of their latus - rectums will always pass through a fixed point.

Let P be any point on the ellipse 2 2

2 2 1x ya b

with y-coordinate k. Prove that the angle between the

tangent at P and the focal chord through P is 2

1tan .baek

A tangent to 2 2

2 2 1x ya b

cuts the axes in A and B and touches the ellipse at P in the first quadrant.

What is the equation of this tangent if AP = PB ?

If the tangent at any point on 2 2

2 2 1x ya b

makes an angle with the major axis and an angle with

the focal radius of the point of contact, show that the eccentricity of the ellipse e satisfies cos .cos

e

Prove that the tangent at any point on the ellipse bisects the external angle between the focal radii ofthat point.

TRY YOURSELF - II

Section - 2 TANGENTS · Mathematics / Ellipse As in circles and parabolas, the equation of a...

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